$f\,^{\prime}(x)=-\dfrac{50}{x^3}$ and $f(1)=20$. $f(5) = $
Answer: Finding $f(x)$ We have $f'(x)=-\dfrac{50}{x^3}$ and we want to find $f(x)$ : $\begin{aligned}f(x)& = \int f'(x)\,dx \\\\ & = \int (-\dfrac{50}{x^3})\,dx \\\\ & = {25x^{-2}} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(1)=20$. Here's what we get when we plug in $1$ : $\begin{aligned}f(1)&={25(1)^{-2}} {+ C}\\\\ &={25} {+ C} \end{aligned}$ We are given that this must equal $20$ : $20 = {25} {+ C}$ Solving the equation gives us ${C=-5}$. Finding $f(5)$ Now, we have that $f(x)= {25x^{-2}} {-5}$. Let's find $f(5)$ by plugging in $5$ : $\begin{aligned}f(5)&=25(5)^{-2}-5\\\\ &=-4 \end{aligned}$ The answer $f(5) = -4$